How could I do this problem if I didn't have a calculator available?

Question

This is actually a sample GRE problem: a quantitative comparison between sqrt(79*80*81*82) and 6400. I recognized that 6400 = sqrt(80*80*80*80) and was therefore faced with comparing 79*81*82 to 80*80*80. I got the right answer, but it was only a guess on my part. So if I didn't have a calculator handy, how could I know that 79*80*81*82 was greater than 80*80*80*80?

I know that it is NOT true, for an arbitrary integer x, that (x-1)x(x+1)(x+2) > x^4. (It doesn't, for example, work for 1, or -1, or -2, or 0.)

Answer 1

79*81*82
= (80-1)(80+1)*82
= (80² - 1)*(80+2)
= 80³ + 2*80² - 80 + 2

2*80² -80 + 2 > 0

79*81*82 > 80³

Answer 2

A formula cant be generalized for all cases in your problem.

See this case:

Which is bigger?

(x-1)x(x+1) x^3

using formula (x-1)*(x+1) = x^2-1, above becomes

(x^2-1)x x^3

simplifying,

x^3-x x^3

Can you say which is bigger here? No right. But if you have an idea if X is positive or negative or zero, then you can figure out the answer without calculator. You may have to use some algebraic simplifications though.

In your case, you could have solved the problem with expanding the LHS like above. Hope this helps.

Answer 3

ha. But does it work for a given value?
I'll skip the square root part (which raises the problem of the +/- which ruins the inequality proof)

2-step proof

(n-1)*n*(n+1)*(n+2) > n^4 ?

step one, proving it for a value of n:

for n = 5
4*5*6*7 = 840 > 625 = 5^4

It is true for n = 5

step two, by induction.
For n>5, given that the inequality is true for n, does it remain true for (n+1)?

For n the formula is:
(n-1)*n*(n+1)*(n+2) > n^4

Is it true for (n+1)?
The formula is now:

n*(n+1)*(n+2)*(n+3) > (n+1)^4

multiply out both sides to get

n^4 + 5n^3 + 7n^2 + 5n + 6 > n^4 + 4n^3 + 6n^2 + 4n + 1

subtract (n^4 + 4n^3 + 6n^2 + 4n + 1) from both sides to get:
n^3 +n^2 + n + 5 > 0
which is true for any value of n greater than 5.

Therefore, we have proven that
n*(n+1)*(n+2)*(n+3) > (n+1)^4
is true for all values of n greater than or equal to 5.

This does not mean it can't work below n=5, simply that we have not proven it.

As it runs out, it works from n=2
1*2*3*4 = 24 > 16 = 2^4